Question

pls tell pls telll pls with explanation urgent pls tell​

Answer 1

$\sf{\bold{\green{\underline{\underline{Given}}}}}$

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• Principle = Rs. 3000
• Amount = Rs. 3993
• Time = 3 years

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$\sf{\bold{\green{\underline{\underline{To\:Find}}}}}$

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• Rate = ??

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$\sf{\bold{\green{\underline{\underline{Solution}}}}}$

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$\sf{\red{\boxed{\bold{Amount = P \bigg\lgroup 1 + \dfrac{rate}{100}\bigg\rgroup^{time}}}}}$

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$\sf :\implies\: {\bold{ 3993 = 3000 \bigg\lgroup 1 + \dfrac{rate}{100} \bigg\rgroup^{3} }}$

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$\sf :\implies\: {\bold{ \dfrac{3993}{3000} = \bigg\lgroup 1 + \dfrac{rate}{100} \bigg\rgroup^{3} }}$

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$\sf :\implies\: {\bold{ \dfrac{1331}{1000} = \bigg\lgroup 1 + \dfrac{rate}{100} \bigg\rgroup^{3} }}$

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$\sf :\implies\: {\bold{ \bigg\lgroup \dfrac{11}{10}\bigg\rgroup ^3 = \bigg\lgroup 1 + \dfrac{rate}{100} \bigg\rgroup^{3} }}$

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• Acc. to the law of exponent same powers implies same bases

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$\sf :\implies\: {\bold{\dfrac{11}{10} = 1 + \dfrac{rate}{100} }}$

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$\sf :\implies\: {\bold{\dfrac{11}{10} - 1 = \dfrac{rate}{100} }}$

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$\sf :\implies\: {\bold{\dfrac{11 - 10 }{10} = \dfrac{rate}{100} }}$

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$\sf :\implies\: {\bold{\dfrac{1}{10} = \dfrac{rate}{100} }}$

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$\sf :\implies\: {\bold{rate = \dfrac{100}{10} }}$

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$\sf :\implies\: {\bold{ rate = 10\% }}$

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$\sf{\bold{\green{\underline{\underline{Answer}}}}}$

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• Rs. 3000 will amount to Rs. 3993 in 3 years at 10% per annum compounded annually

Answer 2

Given :-

• principal ( p ) = rs 3000

• amount ( a ) = rs 3993

• time ( t ) or ( n )= 3 years

To Find :

• rate of the compound interest ( r )

Solution :

• NOW , we know that interest is being compound

So formula for compound interest :

$: \implies \boxed{ \sf{a = p\bigg(1 + \dfrac{r}{100} \bigg) {}^{n} }}$

where ,

• a = amount .

• p = principal

• r = rate

• n = time ( in years )

Substitute all values :

$:\implies \sf \: \: \: \: \sf{3993 = 3000\bigg(1 + \dfrac{r}{100} \bigg) {}^{3} } \\ \\ \\ </p><p>:\implies \sf \: \: \: \: \sf{\bigg(1 + \dfrac{r}{100} \bigg) {}^{3} = \dfrac{3993}{3000} } \\ \\ \\ </p><p>:\implies \sf \: \: \: \: \sf{\bigg(1 + \dfrac{r}{100} \bigg) {}^{3} = \dfrac{1331}{1000} } \\ \\ \\ </p><p>:\implies \sf \: \: \: \: \sf{ \sqrt[3]{ \bigg(1 + \dfrac{r}{100} \bigg) {}^{3}} = \sqrt[3]{\dfrac{3993}{3000} } }$

$:\implies \: \: \: \: \sf{ \bigg(1 + \dfrac{r}{100} \bigg) = \dfrac{11}{10}} \\ \\ \\ </p><p> :\implies \: \: \: \: \sf{ \dfrac{r}{100} = \dfrac{11}{10} - 1} \\ \\ \\ </p><p> :\implies \: \: \: \: \sf{ \dfrac{r}{100} = \dfrac{11 - 10}{10} } \\ \\ \\ </p><p> :\implies \: \: \: \: \sf{ \dfrac{r}{100} = \dfrac{1}{10} } \\ \\ \\ </p><p> :\implies \: \: \: \: \sf{ r= \dfrac{100}{10} } \\ \\ \\ </p><p> :\implies \: \: \: \: \sf{ r = 10}$

• The rate is 10 %

More to know :

Compound Interest Definition :

• Compound interest is the interest calculated on the principal and the interest accumulated over the previous period. It is different from the simple interest where interest is not added to the principal while calculating the interest during the next period.

• Compound interest finds its usage in most of the transactions in the banking and finance sectors and also in other areas as well.

Some of its applications are:

• Increase or decrease in population The growth of bacteria.

• Rise or Depreciation in the value of an item.